Timeline for Power Supply Design and Heating issue of NodeMcu v1.0 ESP-12E
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 26 at 14:35 | comment | added | winny | @arjunkumar There is some cooldown timer for users with low reputation. 24 hours or something similar. It should appear soon. | |
| Sep 26 at 13:07 | comment | added | arjun kumar | @winny, its not allowing me to accept my own answer. thats why I changed the question name | |
| Sep 26 at 13:05 | history | edited | arjun kumar | CC BY-SA 4.0 |
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| Sep 26 at 11:24 | comment | added | winny | @arjunkumar Please don't change the title. If your question is answered, mark the answer as accepted. You can accept your own answer. | |
| Sep 26 at 10:51 | history | edited | arjun kumar | CC BY-SA 4.0 |
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| Sep 26 at 10:35 | answer | added | arjun kumar | timeline score: 0 | |
| Sep 25 at 18:18 | comment | added | arjun kumar | @Aaron, i am using a relay to detect power failure. So by using a 7805 i want to switch on a relay and when relay switches on, then i will make output of 7806 (which has same 9v,1A DC input) to power the Nodemcu v1.0 . when the power failure occurs, the relay will connect to the battery and Nodemcu v 1.0 will not have any power interruption. | |
| Sep 25 at 18:13 | comment | added | arjun kumar | @Aron , to fix the problem, should i have to choose a LDO and also a different power supply ? is there any option to use another ready to use circuit which can take 9V-1A DC input and can produce 6v supply for Nodemcu v1.0 ESP-12E without causing heating issues ?? because i want to use this same 9v,1A DC supply to power a 5V relay. if i change the power supply ,then 5v relay will have a problem?? | |
| Sep 25 at 17:55 | comment | added | Aaron | Almost. The Power is (Vin-Vout) * I. So (9V-5V) *100mA = 400mW. This is enough to make that little part toasty from the start! You can lower your Vin to lower the power dissipated, but the 7805 needs at least 2V of overhead to operate, or Vin=7V. You can get an LDO that needs less than a volt of overhead and drop your Vin, and that'll drop your wasted power a lot. | |
| Sep 25 at 17:50 | comment | added | arjun kumar | @Aron, the output voltage stood at 5v for 7805 regulator. so the power consumed is 5V*100mA = 0.5Watt | |
| Sep 25 at 17:36 | comment | added | Aaron | The current increasing can be normal, it just depends on what the code is having the module do. ie. turn on bluetooth, turn on wifi, change oscillator freq. etc. You measured 100mA at the start, what is that power? Is it enough to make the regulator hot? | |
| Sep 25 at 17:32 | comment | added | arjun kumar | @Aaron,i am not able to understand your question. Just after switching on the power, the current consumption is increasing linearly for every 10 seconds, so where exactly i can do the measurement ??? | |
| Sep 25 at 17:20 | comment | added | Aaron | Did you calculate the Watts being dissipated in the 7805 at 100mA? | |
| Sep 25 at 16:18 | history | edited | arjun kumar | CC BY-SA 4.0 |
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| Sep 25 at 16:13 | history | asked | arjun kumar | CC BY-SA 4.0 |